Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x-4y &= -3 \\ 6x+9y &= -2\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $9y = -6x-2$ Divide both sides by $9$ to isolate $y$ $y = {-\dfrac{2}{3}x - \dfrac{2}{9}}$ Substitute this expression for $y$ in the first equation. $2x-4({-\dfrac{2}{3}x - \dfrac{2}{9}}) = -3$ $2x + \dfrac{8}{3}x + \dfrac{8}{9} = -3$ Simplify by combining terms, then solve for $x$ $\dfrac{14}{3}x + \dfrac{8}{9} = -3$ $\dfrac{14}{3}x = -\dfrac{35}{9}$ $x = -\dfrac{5}{6}$ Substitute $-\dfrac{5}{6}$ for $x$ back into the top equation. $2( -\dfrac{5}{6})-4y = -3$ $-\dfrac{5}{3}-4y = -3$ $-4y = -\dfrac{4}{3}$ $y = \dfrac{1}{3}$ The solution is $\enspace x = -\dfrac{5}{6}, \enspace y = \dfrac{1}{3}$.